in your case you just need to show that f(u,v) = (R + r cosv)cosu is smooth, since the parameterization (u,v) is already smooth.
Hello,
I want to show that this function is smooth:
T is the Torus defined as: and R,r>0 constant.
I don't know how f could be differentiated. My Problem is, that f is not in the standard form f(x,y,z)=x. it is something like f((R+r.....))=(R+r*cosu)cosv.
I'm a little bit confused. Is here any trick or something
Thank you for your help
regards
Hello,
i have shown, that f(x,y,z)=x is smooth (the projection on first coord.) and we know that our parametrization is also smooth, i.e.
But i'm not really sure, why this is enough for being smooth related to f.
Anyway:
Now i want to show, that our function f is a morse function.
But i can't see the critical points of f??
if i differentiate f(x,y,z)=x i get Df=(1,0,0): this is a surjective function isn't it? so there are no critical points.
Can you explain me, what is wrong in my way of thinking?
Thanks
Regards
a "smooth structure" needs to be defined to define smooth functions, just as a manifold does. So I said "in your case", that is, if we don't involve the abstract definition of a manifold, we just define a "smooth structure" on a surface as its smooth parametrization, that is, a smooth map from an area D of R^2 to R^3.
Then a smooth function from a surface is defined to be the smooth function from its parameter domain D.
For your new question, f DOES have critical points on T. You can see this from the expression f(u,v) = (R + r cosv)cosu, or just think that f is defined on a compact surface it must have maximum and minimum points, those points being critical points.
The problem of your way is, you need to "restrict" f to T. Suppose the natural inclusion map from T to R^3 is i, that is, i(x,y,z)=(x,y,z), then the gradient of f restricted to T is i*(df), that is, the pull back of df by i.